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Finding an upper bound on composite $C^1$ functions
Constructing sequence of functionsUpper bound on the inverse of a Grammian matrixLeast Upper Bound Property Implies Greatest Lower Bound PropertyVector Euclidean norm upper bound by his coordinates average.a problem using upper boundsCan n-th derivative of Schwartz function go to $infty$ as $ntoinfty$?Mathematical Analysis by Walter Rudin, Theorem 1.11: Upper/Lower Bounds and Supremum/Infimum.Motivation to define a norm for matrix whereas it's not bounded.Real Analysis - upper and lower boundsQuestion about the range of a norm preserving extension.
$begingroup$
The conditions for the problem I am currently tackling is as follows:
Let $T, M > 0$. Define :
$X_T := $
where $||u||_T = ^textsup_t in (0,T) [|u(t)| +|u'(t)|]$ is the norm of $C^1 ([0,T])$
Suppose that $f : mathbbR rightarrow mathbbR$ belongs to $C^1 (mathbbR)$ and that $f(0) = 0$.
For $a in mathbbR$ and $u in C^1 ([0,T])$ define:
$psi(u)(t) = a + int^t_0 f(u(s)) textds$
Small note, not sure if this is relevant, but I have previously shown that $X_T$ as defined above is a complete metric space under the above norm.
The question for this problem is as follows:
Find $M,T > 0$ (or a suitable condition on $M,T > 0$) such that $psi$ defines a map from $X_T$ to itself.
I have found a minimum value $M$ must take in terms of $a, T,$ and $u$ for a given $u in X_T$ but I can't extend this to a general value that works for all $u$.
I would like to be able to find an upper bound for $^textsup_t in (0,T) |f(u(t))|$ over all $u in X_T$ but does such a bound exist? I know there is such a bound for each $u$ individually, but I don't think our conditions on $f$ are strict enough that $f$ is bounded on $mathbbR$. We would need $f$ to be bounded on $mathbbR$ for this to work right?
Can anyone confirm that my hunch is correct, and if so suggest a different approach I might take? Thank you.
functional-analysis functions norm normed-spaces upper-lower-bounds
asked Mar 7 at 16:43
David HughesDavid Hughes
1607
$endgroup$
add a comment |
$begingroup$
The conditions for the problem I am currently tackling is as follows:
Let $T, M > 0$. Define :
$X_T := $
where $||u||_T = ^textsup_t in (0,T) [|u(t)| +|u'(t)|]$ is the norm of $C^1 ([0,T])$
Suppose that $f : mathbbR rightarrow mathbbR$ belongs to $C^1 (mathbbR)$ and that $f(0) = 0$.
For $a in mathbbR$ and $u in C^1 ([0,T])$ define:
$psi(u)(t) = a + int^t_0 f(u(s)) textds$
Small note, not sure if this is relevant, but I have previously shown that $X_T$ as defined above is a complete metric space under the above norm.
The question for this problem is as follows:
Find $M,T > 0$ (or a suitable condition on $M,T > 0$) such that $psi$ defines a map from $X_T$ to itself.
I have found a minimum value $M$ must take in terms of $a, T,$ and $u$ for a given $u in X_T$ but I can't extend this to a general value that works for all $u$.
I would like to be able to find an upper bound for $^textsup_t in (0,T) |f(u(t))|$ over all $u in X_T$ but does such a bound exist? I know there is such a bound for each $u$ individually, but I don't think our conditions on $f$ are strict enough that $f$ is bounded on $mathbbR$. We would need $f$ to be bounded on $mathbbR$ for this to work right?
Can anyone confirm that my hunch is correct, and if so suggest a different approach I might take? Thank you.
functional-analysis functions norm normed-spaces upper-lower-bounds
asked Mar 7 at 16:43
David HughesDavid Hughes
1607
$endgroup$
$begingroup$
Ah, I feel foolish now. The key is probably the fact that the norm of $u$ is always less than or equal to $2M$. I'm working on it now...
$endgroup$
– David Hughes
Mar 7 at 16:48
add a comment |
$begingroup$
The conditions for the problem I am currently tackling is as follows:
Let $T, M > 0$. Define :
$X_T := $
where $||u||_T = ^textsup_t in (0,T) [|u(t)| +|u'(t)|]$ is the norm of $C^1 ([0,T])$
Suppose that $f : mathbbR rightarrow mathbbR$ belongs to $C^1 (mathbbR)$ and that $f(0) = 0$.
For $a in mathbbR$ and $u in C^1 ([0,T])$ define:
$psi(u)(t) = a + int^t_0 f(u(s)) textds$
Small note, not sure if this is relevant, but I have previously shown that $X_T$ as defined above is a complete metric space under the above norm.
The question for this problem is as follows:
Find $M,T > 0$ (or a suitable condition on $M,T > 0$) such that $psi$ defines a map from $X_T$ to itself.
I have found a minimum value $M$ must take in terms of $a, T,$ and $u$ for a given $u in X_T$ but I can't extend this to a general value that works for all $u$.
I would like to be able to find an upper bound for $^textsup_t in (0,T) |f(u(t))|$ over all $u in X_T$ but does such a bound exist? I know there is such a bound for each $u$ individually, but I don't think our conditions on $f$ are strict enough that $f$ is bounded on $mathbbR$. We would need $f$ to be bounded on $mathbbR$ for this to work right?
Can anyone confirm that my hunch is correct, and if so suggest a different approach I might take? Thank you.
functional-analysis functions norm normed-spaces upper-lower-bounds
asked Mar 7 at 16:43
David HughesDavid Hughes
1607
$endgroup$
The conditions for the problem I am currently tackling is as follows:
Let $T, M > 0$. Define :
$X_T := $
where $||u||_T = ^textsup_t in (0,T) [|u(t)| +|u'(t)|]$ is the norm of $C^1 ([0,T])$
Suppose that $f : mathbbR rightarrow mathbbR$ belongs to $C^1 (mathbbR)$ and that $f(0) = 0$.
For $a in mathbbR$ and $u in C^1 ([0,T])$ define:
$psi(u)(t) = a + int^t_0 f(u(s)) textds$
Small note, not sure if this is relevant, but I have previously shown that $X_T$ as defined above is a complete metric space under the above norm.
The question for this problem is as follows:
Find $M,T > 0$ (or a suitable condition on $M,T > 0$) such that $psi$ defines a map from $X_T$ to itself.
I have found a minimum value $M$ must take in terms of $a, T,$ and $u$ for a given $u in X_T$ but I can't extend this to a general value that works for all $u$.
I would like to be able to find an upper bound for $^textsup_t in (0,T) |f(u(t))|$ over all $u in X_T$ but does such a bound exist? I know there is such a bound for each $u$ individually, but I don't think our conditions on $f$ are strict enough that $f$ is bounded on $mathbbR$. We would need $f$ to be bounded on $mathbbR$ for this to work right?
Can anyone confirm that my hunch is correct, and if so suggest a different approach I might take? Thank you.
functional-analysis functions norm normed-spaces upper-lower-bounds
functional-analysis functions norm normed-spaces upper-lower-bounds
asked Mar 7 at 16:43
David HughesDavid Hughes
1607
asked Mar 7 at 16:43
David HughesDavid Hughes
1607
asked Mar 7 at 16:43
David HughesDavid Hughes
1607
asked Mar 7 at 16:43
David HughesDavid Hughes
1607
asked Mar 7 at 16:43
David HughesDavid Hughes
1607
1607
$begingroup$
Ah, I feel foolish now. The key is probably the fact that the norm of $u$ is always less than or equal to $2M$. I'm working on it now...
$endgroup$
– David Hughes
Mar 7 at 16:48
add a comment |
$begingroup$
Ah, I feel foolish now. The key is probably the fact that the norm of $u$ is always less than or equal to $2M$. I'm working on it now...
$endgroup$
– David Hughes
Mar 7 at 16:48
$begingroup$
Ah, I feel foolish now. The key is probably the fact that the norm of $u$ is always less than or equal to $2M$. I'm working on it now...
$endgroup$
– David Hughes
Mar 7 at 16:48
$begingroup$
Ah, I feel foolish now. The key is probably the fact that the norm of $u$ is always less than or equal to $2M$. I'm working on it now...
$endgroup$
– David Hughes
Mar 7 at 16:48
add a comment |
1 Answer
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$begingroup$
It turns out that my professor made a mistake in setting this question. It does not necessarily have a solution unless we are given a bound $|f| leq 1$ on some $(−2M,2M)$
answered Mar 21 at 18:42
David HughesDavid Hughes
1607
$endgroup$
add a comment |
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$begingroup$
It turns out that my professor made a mistake in setting this question. It does not necessarily have a solution unless we are given a bound $|f| leq 1$ on some $(−2M,2M)$
answered Mar 21 at 18:42
David HughesDavid Hughes
1607
$endgroup$
add a comment |
$begingroup$
It turns out that my professor made a mistake in setting this question. It does not necessarily have a solution unless we are given a bound $|f| leq 1$ on some $(−2M,2M)$
answered Mar 21 at 18:42
David HughesDavid Hughes
1607
$endgroup$
add a comment |
$begingroup$
It turns out that my professor made a mistake in setting this question. It does not necessarily have a solution unless we are given a bound $|f| leq 1$ on some $(−2M,2M)$
answered Mar 21 at 18:42
David HughesDavid Hughes
1607
$endgroup$
It turns out that my professor made a mistake in setting this question. It does not necessarily have a solution unless we are given a bound $|f| leq 1$ on some $(−2M,2M)$
answered Mar 21 at 18:42
David HughesDavid Hughes
1607
answered Mar 21 at 18:42
David HughesDavid Hughes
1607
answered Mar 21 at 18:42
David HughesDavid Hughes
1607
answered Mar 21 at 18:42
David HughesDavid Hughes
1607
1607
add a comment |
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$begingroup$
Ah, I feel foolish now. The key is probably the fact that the norm of $u$ is always less than or equal to $2M$. I'm working on it now...
$endgroup$
– David Hughes
Mar 7 at 16:48